Question

If a metallic sphere gets cooled from ${62}^{\circ }C$ to ${50}^{\circ }C$ in 10 min and in the next 10 min gets cooled to ${42}^{\circ }C$, then the temperature of the surroundings is

• A. ${30}^{\circ }C$
• B. ${36}^{\circ }C$
• C. ${26}^{\circ }C$
• D. ${20}^{\circ }C$

Answer 1

The correct option is C ${26}^{\circ }C$

We know that rate of cooling
$\frac{T_1-T_2}{t}=K[\frac{T_1+T_2}{2}-T_0]$
$\therefore \frac{62-50}{10}=K[\frac{62+50}{2}-T_0]$
$\Rightarrow \frac{12}{10}=K[56-T_0]$...(i)
and $\frac{50-42}{10}=K[\frac{50+42}{2}-T_0]$
$\Rightarrow \frac{8}{10}=K[46-T_0]$...(ii)
Solving Eqs. (i) and (ii), we get
$T_0={26}^{\circ }C$